Question

the sum of n terms of two arithmetic progression are ratio 3n-1:3n+1 then find the ratio of their 5th term​

Answer 1

Answer:

16 / 17

Step-by-step explanation:

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Answer 2

The ratio of the 5th term of both the AP is 13:14.

Given:

The ratio of the nth term of two AP is 3n-1:3n+1

To Find:

The ratio of the nth term

Solution:

Let a be the first term of the first A.P and A be the second term of the first A.P and r be the common difference of the first AP. R is the common difference of second AP.

The ratio Sum of n terms of two AP

$\frac{2a+(n-1)d}{2A+(n-1)D} = \frac{3n-1}{3n+1}$

Taking 2 commons out in LHS

$\frac{2(a+\frac{(n-1)}{2} d)}{2(A+\frac{(n-1)}{2}D)} = \frac{3n-1}{3n+1} \\\\\frac{a+\frac{(n-1)}{2} d}{A+\frac{(n-1)}{2}D} = \frac{3n-1}{3n+1}$

To get the 5th term, we have to put

$\frac{n-1}{2} =4\\n-1 = 8\\n=9$

At n =11

$\frac{a+\frac{(9-1)}{2} d}{A+\frac{(9-1)}{2}D} = \frac{3\times9-1}{3\times9+1} \\\\\\\frac{a+4d}{A+4D} = \frac{27-1}{27+1}\\\\\frac{a+4d}{A+4D} = \frac{26}{28} \\\\\frac{a+4d}{A+4D} = \frac{13}{14}$

Therefore, The ratio of the 5th term of both the AP is 13:14.

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Learn More

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